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1. |
Find the value of `(sumsum)_(0leiltjlen) (1+j)(""^(n)C_(i)+""^(n)C_(j))`. |
Answer» Here sum does nto change if we replace `i` by `n-1` and j by `n-j`. By doing so, in fact we are writing the series in the reverse order. Therefore, `S = underset(0leiltjlen)(sumsum)(i+j)(.^(n)C_(i)+.^(n)C_(j))" "(1)` `=underset(0leiltjlen)(sumsum)(n-i+n-j)(.^(n)C_(n-i)+.^(n)C_(n-j))` `=underset(0leiltjlen)(sumsum)(2n(-i+j))(.^(n)C_(i) + .^(n)C_(j))" "(2)` Adding (1) and (2), we have `2S = 2n underset(0leiltjlen)(sumsum)(.^(n)C_(i) +.^(n)C_(j))` or `S = n xx n2^(n) = n^(2)2^(n)` |
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